Show that $V$ is a vector space when $V$ is the set of all continuous functions $\Bbb{S}^1 \to \Bbb{R}$. Let $V$ (which is infinitely dimensional) be the set of all continuous functions $\Bbb{S}^1 \to \Bbb{R}$.
Show that $V$ is a vector space.
Define $\langle-,-\rangle: V\times V\to \Bbb{R}$ by $$\langle f,g\rangle=\dfrac1\pi\int_{-\pi}^\pi f(x)g(x)dx.$$
I am very lost on how to even start this problem. 
Any help would be appreciated. 
 A: You must show that $V$ satisfies the axioms of a (real) vector space.
As Sammy points out in the comments, the inner product is not relevant to showing that $V$ is a vector space.

Added: Let us consider a more general question, for a moment. Suppose $X$ is any set, and that $\Bbb K$ is any field, and let $\Bbb K^X$ denote the set of functions $X\to\Bbb K.$ (In the case that $X=\emptyset,$ then $\Bbb K^X$ is the set whose only element is the empty function.) Given $f,g\in\Bbb K^X$ and $\alpha\in\Bbb K,$ we define addition and scalar multiplication as follows:


*

*For all $x\in X,$ $(f\oplus g)(x):=f(x)+g(x).$

*For all $x\in X,$ $(\alpha f)(x):=\alpha\cdot f(x).$


It can be shown that for any $f,g\in\Bbb K^X$ and any $\alpha\in K,$ we have $f\oplus g,\alpha f\in\Bbb K^X.$ (This is true even when $X=\emptyset,$ vacuously, since there are no $x\in X.$) Since $\Bbb K$ is a field, then one can readily verify that $\Bbb K^X$ (with addition and scalar multiplication defined as above) satisfies all the axioms of a vector space over $\Bbb K$.
Now, in the case that $X$ and $\Bbb K$ have some sort of topological structure on them (so that there is a notion of continuity of functions $\Bbb X\to\Bbb K$), then we can restrict our attention to the subset $V$ of continuous functions $X\to\Bbb K.$ All that will remain to verify is that for $f,g\in V$ and $\alpha\in\Bbb K,$ we have $f\oplus g,\alpha f\in V.$ Once we've done that, it will follow that $V$ (with addition and scalar multiplication defined as above) is likewise a vector space over $\Bbb K.$ Your particular case is simply $X=\Bbb S^1$ and $\Bbb K=\Bbb R.$
